Let $V$ be a $n$dimensional complex vector space. Using Grothendieck's notation, we define the Grassmannian $G(k,V)$ as the space of $k$quotients of $V$ or, equivalently, as $$ G(k,V)=\{ \mathbb P W \subset \mathbb P V: \dim W=k\}. $$ On $G(k,V)$ we have the tautological bundle sequence: $$ 0 \to \mathcal S^\vee \to V \otimes \mathcal O_{G(k,V)} \to \mathcal Q \to 0. $$ We suppose that the following short exact sequence holds: $$ 0 \to \mathcal Q \otimes \mathcal S^\vee \to N \to \wedge^2 \mathcal Q \to 0, $$ where $N$ is another vector bundle on $G(k,V)$. I know that in the case $k=2$ $$ Ext^1(\wedge^2 \mathcal Q, \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V), \mathcal Q \otimes \mathcal S^\vee \otimes (\wedge^2 \mathcal Q)^\vee) $$ become, using $\wedge^2 \mathcal Q=\mathcal O(1)$ and $\mathcal Q(1)=\mathcal Q^\vee$, $$ Ext^1(\mathcal O(1), \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V),\mathcal Q^\vee \otimes \mathcal S^\vee)=H^1(G(k,V), \Omega^1)=\mathbb C, $$ hence the short exact sequence do not split. I expect that this happens also in the general case, but I cannot prove it. How can I proceed?
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$\begingroup$ Which short exact sequence do you want to show does not split? You did not define the vector bundle $N$ (as far as I can tell), so how do we know $N$ is not $\mathcal{Q}\otimes \mathcal{S}^\vee \oplus \wedge^2\mathcal{Q}$? $\endgroup$– Yosemite StanCommented Sep 1, 2021 at 15:25

$\begingroup$ You right! $N$ is the normal bundle of the Grassmannian $G(k,V)$ into an orthogonal Grassmannian $OG(k,V \oplus V')$. I don't know an explicit description of $N$, I only know that it is the middle term of this short exact sequence $\endgroup$– BobechCommented Sep 2, 2021 at 7:48

$\begingroup$ It seems likely that your sequence is given by taking $N=\wedge^2 (V\otimes \mathcal{O})/\wedge^2(\mathcal{S}^\vee)$. Can you prove that? This bundle would give rise to such an exact sequence. $\endgroup$– Yosemite StanCommented Sep 2, 2021 at 12:18

$\begingroup$ I don't understand why it should be such a bundle, even on the fiber level. By the way, I have to understand what is $Ext^1$, because it parametrize the extensions: if it is zero, then $N$ has to the direct sum. My idea is that there is only one extension as in the case $k=2$, that is an Atiyah extension $\endgroup$– BobechCommented Sep 2, 2021 at 14:39
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$\def\CC{\mathbb{C}}$A splitting would be a global map $f : G(k,n) \times \CC^n \to \CC^n$ such that $f(L,v) \in L$ for all $L \in G(k,n)$ and $v \in \CC^n$. But, since $G(k,n)$ is projective and connected and the target $\CC^n$ is affine, any map $f : G(k,n) \times \CC^n \to \CC^n$ is constant on each $G(k,n) \times \{ v \}$. So we would have a map $f : \CC^n \to \CC^n$ such that $f(v) \in L$ for each $v \in \CC^n$ and each $L$ in $G(k,n)$. The only such map is the zero map.